Problem 1 :
y = rx^{2} + s
y = -2
In the system of equations above, r and s are constants. For which of the following values of r and s does the system have exactly two real solutions ?
A) r = -2, s = -1 B) r = -1, s = -2
C) r = 2, s = - 2 D) r = 3, s = 1
Solution :
y = rx^{2} + s ----(1)
y = -2----(2)
Applying (2) in (1), we get
-2 = rx^{2} + s
Option A :
When r = -2, s = -1
-2 = -2x^{2} + (-1)
-2 + 1 = -2x^{2}
-1 = -2x^{2}
1/2 = x^{2}
x = ±√(1/2)
Option B :
When r = -1, s = -2
-1 = -2x^{2} + (-2)
-1 + 2 = -2x^{2}
1 = -2x^{2}
-1/2 = x^{2}
Not defined.
Hence option A is correct.
Problem 2 :
x^{2} - y^{2} = 48
x + y = 12
If (x, y) is the solution to the system of equations above, what is the value of xy ?
A) 28 B) 32 C) 45 D) 64
Solution :
x^{2} - y^{2} = 48 ----(1)
x + y = 12 ----(2)
y = 12 - x
x^{2} - y^{2} = 48
x^{2} - (12 - x)^{2} = 48
x^{2} - (144 - 24x + x^{2}) = 48
x^{2} - 144 + 24x - x^{2} = 48
24x = 48 + 144
24x = 192
x = 192/24
x = 8
Applying the value of x in (2), we get
y = 12 - 8
y = 4
xy = 8(4) = 32
So, the answer is 32.
Problem 3 :
y = -1.5
y = x^{2} + 8x + a
In the system of equations a is a positive a constant. The system has exactly one distinct real solution. What is the value of a ?
Solution :
y = -1.5 ---(1)
y = x^{2} + 8x + a ---(2)
Applying the value of y in (2), we get
-1.5 = x^{2} + 8x + a
x^{2} + 8x + a + 1.5 = 0
Finding nature of roots :
a = 1, b = 8 and c = a + 1.5
b^{2} - 4ac = 0
8^{2} - 4(1)(a + 1.5) = 0
64 - 4 (a + 1.5) = 0
-4a - 6 = -64
-4a = -64 + 6
-4a = -58
a = 29/2
So, the value of a is 29/2.
Problem 4 :
y = 2x^{2} - 21x + 64
y = 3x + a
In the system of equation, a is a constant. The graphs of the equation in the system intersect at exactly one point (x,y) in the xy plane. What is the value of xy ?
Solution :
y = 2x^{2} - 21x + 64 ----(1)
y = 3x + a ----(2)
Applying (2) in (1), we get
3x + a = 2x^{2} - 21x + 64
2x^{2} - 21x - 3x + 64 - a = 0
2x^{2} - 24x + 64 - a = 0
Since it has a point of intersection at (x, y), the above quadratic equation will have real roots.
Finding nature of roots :
a = 2, b = -24 and c = 64 - a
b^{2} - 4ac = 0
(-24)^{2} - 4(2)(64 - a) = 0
576 - 8(64-a) = 0
576 - 512 + 8a = 0
64 + 8a = 0
8a = -64
a = -64/8
a = -8
y = 3x - 8
Applying the value of y,
3x - 8 = 2x^{2} - 21x + 64
2x^{2} - 21x + 64 - 3x + 8 = 0
2x^{2} - 24x + 72 = 0
x^{2} - 12x + 36 = 0
(x - 6)(x - 6) = 0
x = 6 and x = 6
y = 3(6) - 8
y = 10
xy = 6(10) ==> 60
So, the value of xy is 60.
Problem 5 :
x^{2} + 2xy + y^{2} = 25
x - y = 7
If (x, y) is a solution to the system of equations above, what is one possible value of x ?
Solution :
x^{2} + 2xy + y^{2} = 25 ---(1)
x - y = 7 ----(2)
x = 7 + y
Applying the value of x in (1), we get
(7 + y)^{2} + 2xy + y^{2} = 25
49 + 14y + y^{2 }+ 2xy + y^{2} = 25
2y^{2}+ 14y^{ }+ 2xy + 49 - 25 = 0
2y^{2}+ 14y^{ }+ 2xy + 24 = 0
It cannot be solved.
Trying in another way :
(a + b)^{2} = a^{2} + 2ab + b^{2}
(a - b)^{2} = a^{2} - 2ab + b^{2}
a^{2} + b^{2 }= (a - b)^{2 }+ 2ab
x^{2} + y^{2 }= (x - y)^{2 }+ 2xy
Replacing the above in (1), we get
x^{2} + y^{2}+ 2xy = 25
(x - y)^{2 }+ 2xy + 2xy = 25
7^{2 }+ 4xy = 25
4xy = 25 - 49
4xy = -24
xy = -24/4
xy = -6
So, the value of xy is -6.
Problem 6 :
(1/2) x^{2} - 3y^{2} = 55
x = -4y
If (x, y) is a solution to the system of equations above, what is the value of y^{2} ?
Solution :
(1/2) x^{2} - 3y^{2} = 55 -----(1)
x = -4y -----(2)
Applying the value of x from (2) in (1)
(1/2) (-4y)^{2} - 3y^{2} = 55
(1/2) (16y)^{2} - 3y^{2} = 55
8y^{2}- 3y^{2} = 55
5y^{2} = 55
y^{2} = 55/5
y^{2} = 11
So, the value of y^{2} is 11.
Problem 7 :
x - y + 2 = 0
(x + 2)^{2} - 7(x + 2) + 25 = 4y - 5
If (x, y) is a solution to the system of equations above, what is one possible value of x ?
Solution :
x - y + 2 = 0 ---(1)
(x + 2)^{2} - 7(x + 2) + 25 = 4y - 5 ----(2)
From (1),
y = x + 2
Applying the value of y in (2), we get
y^{2} - 7y + 25 = 4y - 5
y^{2} - 7y - 4y + 25 + 5 = 0
y^{2} - 11y + 30 = 0
(y - 6)(y - 5) = 0
y = 6 and y = 5
Applying y = 6, we get x = 6 - 2 ==> 4
Applying y = 6, we get x = 5 - 2 ==> 3
So, the values of x are 4 and 3.
Problem 8 :
f(x) = x^{2} - 3x
g(x) = 2x + 14
The functions f and g are defined above. For how many values of k is it true that f(k) = g(k) ?
A) one B) None C) Two D) more than two
Solution :
f(x) = x^{2} - 3x
g(x) = 2x + 14
f(k) = g(k)
k^{2} - 3k = 2k + 14
k^{2} - 3k - 2k - 14 = 0
k^{2} - 5k - 14 = 0
(k - 7) (k + 2) = 0
k = -2 and k = 7
So, it will have two solutions.
Problem 9 :
If (x + y)^{2} - (x - y)^{2} = 60 and x and y are positive integers, which of the following could be the value of x + y?
A) 6 B) 8 C) 10 D) 12
Solution :
(x + y)^{2} - (x - y)^{2} = 60
x^{2} + y^{2} + 2xy - (x^{2} + y^{2} - 2xy) = 60
x^{2} + y^{2} + 2xy - x^{2} - y^{2} + 2xy = 60
4xy = 60
xy = 60/4
xy = 15
The possible values of x and y are
x = 3, y = 5 or x = 1 and y = 15
x + y = 3 + 5 ==> 8
(or)
x + y = 1 + 15 = 16
So, option B is correct.
Problem 10 :
y = 3x - 1
y = (x + 1)^{2}
The system of equations above has how many solutions ?
A) 0 B) 1 C) 2 D) infinitely many
Solution :
y = 3x - 1 ----(1)
y = (x + 1)^{2 }----(2)
(1) = (2)
3x - 1 = (x + 1)^{2 }
3x - 1 = x^{2} + 2x + 1
x^{2} + 2x + 1 - 3x + 1 = 0
x^{2} - x + 2 = 0
(x + 2)(x - 1) = 0
x = -2 and x = 1
So, the system has two solutions, option C is correct.
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